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3 years ago

What is the difference between Nuclear Fission and Nuclear Fusion? Also any examples of them

Chemistry

1 answer:

stealth61 [152]3 years ago

6 0

Answer:

Hiya there!

Explanation:

Both fission and fusion are nuclear reactions that produce energy, but the applications are not the same. Fission is the splitting of a heavy, unstable nucleus into two lighter nuclei, and fusion is the process where two light nuclei combine together releasing vast amounts of energy.

For example, uranium can fission to yield strontium and krypton. Fusion joins atomic nuclei together. The element formed has more neutrons or more protons than that of the starting material. For example, hydrogen and hydrogen can fuse to form helium.

Hope this helped! :D

Credit sourced from "nuclear.duke-energy.com, thoughtco.com"

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Most metals are solid at room temperature, malleable, ductile, good conductors of heat and electricity, and react in acids to pr

Evgen [1.6K]

Answer:

The answer to your question is: "react with acids to produce hydrogen gas"

Explanation:

Chemicals properties of matter: are properties that can be measured if there is a chemical change or chemical reaction.

Examples of chemical reactions:

Reactivity

Toxicity

Coordination number

Flammability

Of the properties listed, the only that implies a reaction is "react with acids to produce hydrogen gas"

7 0

3 years ago

Read 2 more answers

Balance this equation. Not all spots have to have a number.

Lisa [10]

2Na+2H2O----2NaOH+H2

3 0

3 years ago

Americans combined drive about 4.0 x 109 kilometers a day and get an average of 20 miles per gallon of gasoline. For each kilogr

Nimfa-mama [501]

Answer:

$303,882.84649 kg\times 3=9.12\times 10^5 kg$ of carbon dioxide gas.

Explanation:

Average distance covered by Americans in a day= $4.0\times 10^9 km$

1 day = 24 × 60 min = 1,440 min

Average distance covered by Americans in a minute= $\frac{4.0\times 10^9 km}{1,440}=2,777,777.78 km$

Average mileage of the car = 20 miles/gal = 32.18 km/gal

1 mile = 1.609 km

20 miles = 20 × 1.609 km = 32.18 km

Volume of gasoline used in minute = $\frac{2,777,777.78 km}{32.18 km/gal}$

$V=86,320.00 gal$

$V=86,320.00\times 3.7854 L$

(1 L = 1000 mL)

$V=86,320.00\times 3.7854 \times 1000 mL=326,755,748.91 mL$

Mass of 86,320.00 gallons of gasoline = m

Density of the gasoline = d = $0.93 g/cm^3=0.93 g/mL$

$1 mL= 1 cm^3$

$m=d\times V=0.93 g/mL\times 326,755,748.91 mL$

$m=303,882,846.49 g=303,882.84649 kg$

1 kilogram of gasoline gives 3 kg of carbon dioxde gas .

Then 303,882.84649 kg of gasoline will give :

$303,882.84649 kg\times 3=9.12\times 10^5 kg$ of carbon dioxide gas.

8 0

3 years ago

How many moles of S are in 35.4 g of (C3H5)2S?

alexandr402 [8]

0.310 moles First, look up the atomic weights of the elements involved. Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight sulfur = 32.065 Molar mass (C3H5)2S = 6 * 12.0107 + 10 * 1.00794 + 32.065 = 114.2086 g/mol Moles (C3H5)2S = 35.4 g / 114.2086 g/mol = 0.309959145 mol Since there's just one sulfur atom per (C3H5)2S molecule, the number of moles of sulfur will match the number of moles of (C3H5)2S which is 0.310 when rounded to 3 significant digits.

6 0

3 years ago

An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t

topjm [15]

Answer:

$\boxed{3}$

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

$\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )$

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

$\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}$

Data:

$n_{f} = 2$

λ = 657 nm

Calculation:

$\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}$

7 0

3 years ago